Prove Chi Square is Stochastically Increasing

  • #1
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Homework Statement


Prove that the [itex]X^2[/itex] distribution is stochastically increasing in its degrees of freedom; that is if [itex]p>q[/itex], then for any [itex]a[/itex], [itex]P(X^2_{p} > a) \geq P(X^2_{q} > a)[/itex], with strict inequality for some [itex]a[/itex].

Homework Equations


1.[itex](n-1)S^2/\sigma^2 \sim X^2_{n-1}[/itex]
2.The Chi squared(p) pdf is
[itex]f(x|p)= \frac 1{\Gamma(p/2) 2^{p/2}}x^{(p/2) - 1}e^{-x/2} [/itex]

The Attempt at a Solution


Since [itex]p>q[/itex], this implies [itex]\forall a, \frac{\sigma^2 a}{p}< \frac{\sigma^2 a}{q} [/itex].
Also, [itex]X^2_{k} \sim kS^2/\sigma^2 [/itex].

Therefore [itex] \forall a, P(X^2_{p}>a) = P(S^2 > \sigma^2 a/p) \geq P(S^2 > \sigma^2 a/q) = P(X^2_{q}>a) [/itex].

If [itex] a>0[/itex], we observe strict inequality, as the support of [itex]S^2[/itex] is [itex][0,\infty) [/itex]...

What do you think? If I am going in the wrong direction, please steer me in the right one.
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Prove that the [itex]X^2[/itex] distribution is stochastically increasing in its degrees of freedom; that is if [itex]p>q[/itex], then for any [itex]a[/itex], [itex]P(X^2_{p} > a) \geq P(X^2_{q} > a)[/itex], with strict inequality for some [itex]a[/itex].

Homework Equations


1.[itex](n-1)S^2/\sigma^2 \sim X^2_{n-1}[/itex]
2.The Chi squared(p) pdf is
[itex]f(x|p)= \frac 1{\Gamma(p/2) 2^{p/2}}x^{(p/2) - 1}e^{-x/2} [/itex]

The Attempt at a Solution


Since [itex]p>q[/itex], this implies [itex]\forall a, \frac{\sigma^2 a}{p}< \frac{\sigma^2 a}{q} [/itex].
Also, [itex]X^2_{k} \sim kS^2/\sigma^2 [/itex].

Therefore [itex] \forall a, P(X^2_{p}>a) = P(S^2 > \sigma^2 a/p) \geq P(S^2 > \sigma^2 a/q) = P(X^2_{q}>a) [/itex].

If [itex] a>0[/itex], we observe strict inequality, as the support of [itex]S^2[/itex] is [itex][0,\infty) [/itex]...

What do you think? If I am going in the wrong direction, please steer me in the right one.
I cannot follow your argument. You need two different ##S^2## random variables---one for ##\chi^2_p## and a different one for ##\chi^2_q##. Basically, though, you need to know what ##\chi^2## really is, in simple, intuitive terms: if ##Y_r## has the distribution ##\chi^2_r## then
[tex] Y_r = Z_1^2 + Z_2^2 + \cdots + Z_r^2, [/tex]
where ##Z_1, Z_2, \ldots, Z_r## are iid standard normal random variables.

When looking at stochastic ordering, you are entitled to use a common sample space ##\Omega##, since all that matters is how the distribution functions compare (not, for example, whether the two random variables are independent, or not). We can always "construct" a sample space such that the iid N(0,1) random variables ##Z_1, Z_2, \ldots, Z_p## are functions over ##\Omega## (so their values "observed" on a sample point ##\omega \in \Omega##) are ##Z_1(\omega), Z_2(\omega), \ldots, Z_p(\omega)##. Then
[tex] Y_q(\omega) = \sum_{j=1}^q Z_j(\omega)^2 \; \text{and} \;Y_p(\omega) = \sum_{j=1}^p Z_j(\omega)^2 [/tex]
For ##q < p##, what does that tell you?
 
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  • #3
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6
Based on your hint, here's what I'm thinking...

Well, then, if [itex]Z_{i} \sim N(0,1)[/itex] (and i.i.d.) , and [itex]X^2_{n} = \sum_{i=1}^{n} Z_{i} [/itex], then [itex]X^2_{p} = pZ^2_1 [/itex] and [itex]X^2_{q} = qZ^2_1 [/itex].

Thus for all [itex]a [/itex], [itex] P(X^2_{p} > a) = P(pZ^2_1 > a ) = P(Z^2_1 > \frac {a}{p}) \geq P(Z^2_1 > \frac {a}{q}) = P(X^2_{q} > 1) [/itex], since [itex] p>q [/itex]. If [itex]a>0 [/itex], then there should be strict equality (I think).

Whatdya think?
 
  • #4
Ray Vickson
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Based on your hint, here's what I'm thinking...

Well, then, if [itex]Z_{i} \sim N(0,1)[/itex] (and i.i.d.) , and [itex]X^2_{n} = \sum_{i=1}^{n} Z_{i} [/itex], then [itex]X^2_{p} = pZ^2_1 [/itex] and [itex]X^2_{q} = qZ^2_1 [/itex].

Thus for all [itex]a [/itex], [itex] P(X^2_{p} > a) = P(pZ^2_1 > a ) = P(Z^2_1 > \frac {a}{p}) \geq P(Z^2_1 > \frac {a}{q}) = P(X^2_{q} > 1) [/itex], since [itex] p>q [/itex]. If [itex]a>0 [/itex], then there should be strict equality (I think).

Whatdya think?
I think you are writing down a bunch of material that makes no sense at all. Chi-squared = a sum of squares of iid N(0,1) random variables, not just a sum. I suggest you sit down, relax, and proceed slowly and carefully. Think about the actual definitions, and think about what I wrote in my previous response.

Alternatively, you can try to proceed directly: ##X## (with cdf ##F##) dominates ##Y## (with cdf ##G##) if ##G(x) \geq F(x)## for all ##x##, and ##>## holds for some ##x##. In other words, the stochastically larger random variable has a smaller cdf; that is, for each ##x##, it lies below the other one---so ##F## describes a distribution the lies to the "right" of ##G##. That means that the density functions ##f,g## must satisfy ##\int_0^x [g(t) - f(t)] \geq 0 ## for all ##x > 0##. You have formulas for ##f(t)## and ##g(t)##, so you can try to verify the cumulative ordering. That might be hard to do, so that is why I suggested an alternative approach.
 
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  • #5
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Oops. Should have written [itex]X^2_{n} = \sum^{n}_{i=1} Z^2_{i} [/itex]. I was thinking that since the [itex]Z_{i}[/itex]'s are identically distributed, that I could just sum up their squares, [itex] X^2_1 [/itex]. Why am I wrong here, if they're identically distributed?

I will try the direct method, since [itex]X^2_{p} \sim \Gamma(p/2, 2) [/itex].

Update: Ok, I believe I've found a solid direct proof using the Chi square, Gamma and Poisson distribution. I'm excited, but busy now, so check tomorrow for my reply! And thanks for the help.
 
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  • #6
120
6
Ok, there is a relationship between the Gamma and Poisson distribution. Let [itex] X \sim Gamma(\alpha, \beta) [/itex], then
[itex] P(X \leq x) = P(Y \geq \alpha) [/itex], where [itex] Y \sim Poisson(x/\beta) [/itex].

Now let [itex]p > q [/itex], [itex] p,q \in \mathbb{Z}^{+}[/itex] [itex] a>0 [/itex], then
[itex] P(X^2_{p} > a) < P(X^2_{q} > a)[/itex]
[itex]\Leftrightarrow P(Y > p) < P(Y > q) [/itex]
[itex]\Leftrightarrow 0 < P(Y>q) - P(Y>p) [/itex]
[itex]\Leftrightarrow 0 < \sum^{\infty}_{y=q/2} \frac {e^{-a/2}(a/2)^{y}}{y!} - \sum^{\infty}_{y=p/2} \frac{e^{-a/2}(a/2)^{p} }{y!} [/itex]
[itex] = \sum^{p/2}_{y=q/2} \frac {e^{-a/2} (a/2)^{y} }{y!} [/itex].
The above is positive, proving a strict inequality for some [itex]a [/itex].

If [itex]a \notin \mathbb{R}^{+} [/itex], then
[itex] P(X^2_{p} > a) = 1 = P(X^2_{q} > a)[/itex],
hence [itex] \forall a \in \mathbb{R}, P(X^2_{p} > a) \geq P(X^2_{q} > a) [/itex], if [itex]p>q [/itex].

I think this is right.
 

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